# Solve the system of linear equations using matrices. x+y+z=3 2x+3y+2z=7 3x-4y+z=4

Question
Matrices
Solve the system of linear equations using matrices.
x+y+z=3
2x+3y+2z=7
3x-4y+z=4

2021-02-14
Step 1
Given equation
x+y+z=3
2x+3y+2z=7
3x-4y+z=4
in matrix form
$$A=\begin{bmatrix}1&1&1 \\2&3&2\\3&-4&1 \end{bmatrix}$$
$$B=\begin{bmatrix}3 \\7 \\4 \end{bmatrix}$$
$$X=\begin{bmatrix}x \\y \\z \end{bmatrix}$$
we know that $$A \cdot X=B$$
So, $$X=A^{-1} \cdot B$$
Step 2
Now,
$$X=A^{-1} \cdot B$$
We have to find $$A^{-1}$$
Adjoin the identity matrix onto the right of the original matrix, so that you have A on the left side and the identity matrix on the right side. $$\begin{bmatrix}1&1&1 \\2&3&2\\3&-4&1 \end{bmatrix}=\begin{bmatrix}1&0&0 \\0&1&0\\0&0&1 \end{bmatrix}$$
now we will apply the elementary row method $$R_2-2R_1 \rightarrow R_2$$ (multiply 1 row by 2 and subtract it from 2 row), $$R_3-3R_1 \rightarrow R_3$$ (multiply 1 row by 3 and subtract it from 3 row)
$$\begin{bmatrix}1&1&1 \\2-2\cdot1&3-2\cdot1&2-2\cdot1\\3-3\cdot1&-4-3\cdot1&1-3\cdot1 \end{bmatrix}=\begin{bmatrix}1&0&0 \\0-2\cdot1&1-2\cdot0&0-2\cdot0\\0-3\cdot1&0-3\cdot0&1-3\cdot0 \end{bmatrix}\\ \begin{bmatrix}1&1&1 \\0&1&0\\0&-7&-2 \end{bmatrix}=\begin{bmatrix}1&0&0 \\-2&1&0\\-3&0&1 \end{bmatrix}$$
$$R_1-1R_2 \rightarrow R_1$$ (multiply 2 row by 1 and subtract it from 1 row), $$R_3 + 7 R_2 \rightarrow R_3$$ (multiply 2 row by 7 and add it to 3 row)
$$\begin{bmatrix}1&0&1 \\0&1&0\\0&0&-2 \end{bmatrix}=\begin{bmatrix}3&-1&0 \\-2&1&0\\-17&7&1 \end{bmatrix}$$
$$\frac{R_3}{-2} \rightarrow R_3$$ (divide the 3 row by -2)
$$\begin{bmatrix}1&0&1 \\0&1&0\\0&0&1 \end{bmatrix}=\begin{bmatrix}3&-1&0 \\-2&1&0\\8.5&-3.5&-0.5 \end{bmatrix}$$
$$R_1-1R_3 \rightarrow R_1$$ (multiply 3 row by 1 and subtract it from 1 row)
$$\begin{bmatrix}1&0&0 \\0&1&0\\0&0&1 \end{bmatrix}=\begin{bmatrix}-5.5&2.5&0.5 \\-2&1&0\\8.5&-3.5&-0.5 \end{bmatrix}$$
$$A^{-1}=\begin{bmatrix}-5.5&2.5&0.5 \\-2&1&0\\8.5&-3.5&-0.5 \end{bmatrix}$$
Step 3
Now,
$$X=A^{-1} \cdot B$$
$$X=\begin{pmatrix}-5.5&2.5&0.5 \\-2&1&0\\8.5&-3.5&-0.5 \end{pmatrix}\times \begin{pmatrix}3 \\7 \\4 \end{pmatrix}$$
$$X=\begin{pmatrix}-5.5 \cdot 3&2.5\cdot7&0.5\cdot4 \\-2\cdot3&1\cdot7&0\cdot4\\8.5\cdot3&-3.5\cdot7&-0.5\cdot4 \end{pmatrix}$$
$$X=\begin{pmatrix}3 \\1 \\-1 \end{pmatrix}$$

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